Problem: $A=\left[\begin{array}{rr}-6 &-18 &3 & -25 & -8 \\2 & 9 & -14 & 4 & -12 \\16 &2 &28 & 3 & 7 \\25 &19 &-37 & 44 & 12\end{array}\right]$ $A_{2,3}=$
Solution: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{2,3}$ $A_{{2},{3}}$ is located on row ${2}$ of $A$ : $\left[\begin{array}{rr}-6 &-18 &3 & -25 & -8 \\ {2} & {9} & {-14} & {4} & {-12} \\16 &2 &28 & 3 & 7 \\25 &19 &-37 & 44 & 12\end{array}\right]$ $A_{{2},{3}}$ is also located on column ${3}$ of $A$. $\left[\begin{array}{rr}-6 &-18 &3 & -25 & -8 \\ {2} & {9} & {\text{-14}} & {4} & {-12} \\16 &2 &{28} & 3 & 7 \\25 &19 &-{37} & 44 & 12\end{array}\right]$ Therefore, $A_{{2},{3}}={-14}$. Summary $A_{2,3}=-14$